1549901369-Elements_of_Real_Analysis__Denlinger_

78 Chapter 2 • Sequences

*Theorem 2.3.10 If {xn} is a sequence of nonzero numbers such that

lim IXn+l I = L < 1, then, Xn ----> 0.

n--+oo Xn

Proof. Suppose { Xn} is a sequence such that lim I Xn+^11 = L < 1. Then

n--+ex> Xn

1 - L > 0. Let o =

1

• L. Then o > 0. Since IXn+l I ----> L, :3 no E N 3
  2 Xn

n ~no =? I lx;:1 I-LI < 5

=? L - o < lx;:1 I < L + o L. (_ \

I

=? --Xn+l I < L + --^1 - L = --L +^1 < t' 1. (Why?)

Xn 2 2

=? --<--<l.

I

Xn+ll L + 1

Xn 2

. L+ 1 q
Takmg r = -
2

• , we haven~ no=? lxn+_!I < r1xnl· Thus,

lxno+1 I < rlxno I

lxno~ < r lxno+1 I < l

2

f xno I

: t£ -)0

lxno+kl < rklxnol· (1)

By Theorem 2.3.7, lim rk = 0, since 0 < r < 1. Then (recall that no is

k--+oo

fixed),

Putting this together with Inequality (1), we have

lxno+kl < rklxn 0 I----> 0. (ask----> oo)

Therefore, by the second squeeze principle, lim lxno+k I

k--+oo

n--+oo lim lxnl =^0 (Exercise 10), so n--+oo lim Xn = 0. •

Corollary 2.3.11 For any fixed real number c, lim (en!) = 0.

n--+oo n.

Proof. Exercise 11. •

0. Therefore,

*An asterisk before a theorem, proof, or other item in this chapter indicates that the item is
challenging or can be omitted, especially in a one-semester course.